The Pressure's On

For the summer, I'm staying at my parents' new condo in downtown Northampton.  It's a nice place, but as usual, it's the physics examples that catch my eye.  The shower head is unusually large, and I've noticed that when I turn off the water, some stays trapped inside the head and slowly drips out.  What's interesting is the way it drips out – since the head is tilted slightly, the drips start off coming from the whole head, but gradually stop in the front as the trapped water diminishes.

Dripping water, as opposed to a stream, is caused by surface tension, the extra binding force at the surface of a liquid.  For a drip clinging to a tube, the force of the surface tension is

where d is the diameter of the tube, and γ is a constant that depends on the specific liquid.  To find when the drip falls, we can set this force equal to the forces acting against it – gravity and the pressure in the tube.

where m is the mass of the drip and p is the pressure in the tube.  Notice that these two are the only real variables in the equation; everything else is a constant.

First, we'll consider the pressure, p.  Suppose the shower head is tilted at an angle θ:

We'll assume it's a cylinder with radius r.  Then we can relate the volume of water V to the depth of the water h at a specific point x by integrating over the area of rectangular slices:

It can be a little difficult to see what exactly h(x) looks like, but I found this diagram helpful:

 

We can think of h(x) as a segment of a right triangle such that

where xc is the length of the clipped portion of the triangle.  Then the volume is

or evaluating,

Using this, we can eliminate xc from our equation for h(x):

or putting it in terms of the pressure at a point x

where ρ is the density of water, and g is the acceleration due to gravity.  Simplifying things a bit, we can write

where M is the mass of the trapped water.  Substituting into the force equation found above, we have

Note that

so by integrating over x, we can turn this into a differential equation.

Solving this gives

where C is some constant.  Interestingly, this says that you'll never lose all the trapped water, since the pressure will eventually drop to zero, but a few drops will stay stuck from the surface tension.  Substituting back into the force equation and solving for m gives

Here's what this looks like (y-axis represents drip size, x-axis represents position):

I admit, that was a lot of work for an only mildly interesting result, but they can't all be winners...